# Does this integral $\int_0^\infty \frac{dx}{(1+e^x)(a+x)}$ have a closed form?

Note that $a>0$, thus I'm not sure if we can apply residues here. (For $a=0$ the integral doesn't converge).

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}$$

Despite the simple expression under the integral, I didn't find it in G-R book.

I attempted the most straightforward way - geometric series - since $e^x \geq 1$.

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty (-1)^{k+1} \int_0^\infty \frac{e^{-kx}dx}{a+x}$$

The integrals on the right can be expressed using incomplete gamma function:

$$\int_0^\infty \frac{e^{-kx}dx}{a+x}=\frac{e^{ka}}{k} \Gamma(0,ka)$$

We obtain:

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty (-1)^{k+1} \frac{e^{ka}}{k} \Gamma(0,ka)$$

This is correct, but not very useful, since each incomplete gamma just disguises an integral. It can be computed independently by:

$$\Gamma (0,t)=\cfrac{\exp(-t)}{t+1-\cfrac{1}{t+3-\cfrac{4}{t+5-\cfrac{9}{t+7-\cdots}}}}$$

Thus I suppose we can express the integral as:

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \cfrac{1}{ka+1-\cfrac{1}{ka+3-\cfrac{4}{ka+5-\cfrac{9}{ka+7-\cdots}}}}$$

Amusing, but again, not very useful.

Does this integral have a closed form? If so, what is it and how do we find it?

Edit

Another interesting expression for incomplete gamma from DLMF:

$$\Gamma (0,t)=e^{-t} \sum_{n=0}^\infty \frac{L_n (t)}{n+1}$$

Here $L_n (t)$ are Laguerre polynomials. This series converges extremely slowly (and oscillates), so it seem quite useless for computation, however gives a pretty expression for the integral:

$$\int_0^\infty \frac{dx}{(1+e^x)(a+x)}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \sum_{n=0}^\infty \frac{L_n (ka)}{n+1}$$

Note, that the continued fraction above converges much faster than the series.

Since: $$\mathcal{L}\left(\frac{1}{1+e^x}\right) = \frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\qquad \mathcal{L}^{-1}\left(\frac{1}{a+x}\right)=e^{-as}$$ the original integral equals: $$\int_{0}^{+\infty}\frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds=\frac{1}{2}\int_{0}^{+\infty}\int_{\frac{s+1}{2}}^{\frac{s+2}{2}}\sum_{n\geq 0}\frac{e^{-as}}{(n+u)^2}\,du\,ds$$ and by integration by parts, the LHS just depends on: $$\int_{0}^{+\infty}\left[\log\Gamma\left(\frac{s+2}{2}\right)-\log\Gamma\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds$$ that can be further expanded by exploiting the Weierstrass product for the $\Gamma$ function, leading to a fast-convergent series, whose first term is $-\frac{\gamma}{2a}$.

• What is $\psi$? – GFauxPas Jun 10 '16 at 14:16
• @GFauxPas, digamma function, I presume – Yuriy S Jun 10 '16 at 14:16
• Do I understand correctly that Laplace transform in this case means just using $$\frac{1}{a+x}=\int_0^\infty e^{-(a+x)t}dt$$, reversing order of integration and then solving $$\int_0^\infty \frac{e^{-tx} dx}{1+e^x}$$? – Yuriy S Jun 10 '16 at 14:30
• @YuriyS: absolutely right. – Jack D'Aurizio Jun 10 '16 at 14:31